Question

The point diametrically opposite to the point $P(1,0)$on the circle $x^2+y^2+2x+4y-3=0$ is

• A. $(3,4)$
• B. $(3,-4)$
• C. $(-3,4)$
• D. $(-3,-4)$

Answer 1

The correct option is D

$(-3,-4)$

Find the point opposite to the given point on the circle

Given: the equation of circle is $x^2+y^2+2x+4y-3=0$ and point $P(1,0)$lies on it.

Centre of the given circle be$(-1,-2)$

Let another end of the diameter be $(x,y)$
$\because$ centre is the mid point of the end points diameter.
$\therefore$ By using mid point formula,
$$\begin{gathered} \frac{x+1}{2}=-1,\frac{y+0}{2}=-2 \\ \Rightarrow x=-3,y=-4 \end{gathered}$$

Therefore, the point is$(-3,-4)$

Hence, option (D) is the correct answer.